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Template:Maths rating

page title

Is "del" really the most common name of the operator? I have heard the names "nabla", "grad" and "gradient" before, but never "del". Maybe the page should be renamed to "Nabla" and just mention "del" as an alternative name for the operator? -- Jochen

The formal operator was certainly called "del" in all of my mathematics and physics courses, which also mentioned the (infrequently used) name "nabla operator". The operations it is typically used to construct are called "div", "grad" and "curl". "Del" is one syllable, "nabla" two, so when you have to speak a lot of equations, it's more natural to use "del". It's also easier to say "del squared" than "laplacian of". -- Karada 11:17, 7 Nov 2004 (UTC)
Asking Google for "del operator" gives 1620 hits, asking for "nabla operator" gives 2380. This would indicate that "nabla" is more common. -- J.Voss 13:50, 7 Nov 2004 (UTC)
But, then, most people say "del", not "the del operator," so it's not clear that this is meant. As below, the google fight for "del+math" vs. "nabla+math" shows del+math on top. So I'm not willing to say either is definitive proof. 09:40, 23 November 2006 (UTC)
Just to add to this, the del symbol itself isn't the same as the "grad" operation; you only get the grad operation by "multiplying" a scalar field by del. -- Karada 11:21, 7 Nov 2004 (UTC)

Perhaps it could also be notes that sometimes the symbol is used for . And even the tex editor you use here for math editing uses nabla instead of del. --Jaap 20:00, 18 Apr 2005 (UTC)

It's always been del to me, from calculus to physics at umass. also: phi is 'fee'


But compare --Jochen 10:09, 29 September 2005 (UTC)
I am also in favour of a move of del to nabla. I think "del" is a manifestation of loss of culture (so we should fight against this). There is no sense imho of calling this "del". The symbol is a reversed delta, OK, but then call it "led" or "atled". (Besides the fact that many other things would deserve the WP "Del" page much more than this one...) — MFH:Talk 15:13, 16 March 2006 (UTC)
This article is about the operator, not the symbol. (talk) 17:24, 3 December 2008 (UTC)
Wikipedia isn't a forum for the preservation of culture. Nabla is the symbol, del is the operator, that makes sense, and the rest is nothing to fret over. 09:40, 23 November 2006 (UTC)
PS: if you add Ostrogradsky, it's 27 against 4 - as I said, a problem of culture... (I better don't mention the result for Ostrogradski... ;-) — MFH:Talk 15:44, 16 March 2006 (UTC)

My VNR Concise Encyclopedia of Mathematics calls it the "nabla operator". Arfken & Weber's Mathematical Methods for Physicists calls the symbol "del". So, if you look at actual usage in respectable sources, I don't think there is a good basis for saying that "del" is only the operator and "nabla" is only the symbol. As for the article title, I don't really have an opinion, as long as all the widespread usages are described evenhandedly. —Steven G. Johnson (talk) 18:45, 3 December 2008 (UTC)

(To be honest, I personally would tend to say that neither del nor nabla are operators. Operators are things like gradient, divergence, and curl, which have a well-defined domain and range. Del/nabla is simply a shorthand notation for expressing all of those related operators in terms of the same mnemonic symbol, emphasizing e.g. that divergence is the dual to the gradient operator. —Steven G. Johnson (talk) 18:50, 3 December 2008 (UTC))


Is it best to expand on how to convert del to different coordinate systems here or in the other coordinate systems? This is done very poorly at them moment, from new unit vectors (and relationship to i,j,k), why you get (1/r).(d/d theta) e.t.c. --rex_the_first 18:30, 05 Nov 2005 (GMT)

Yeah, sorry about that; I know it's not incredibly obvious. Sadly, there are easy ways to do it (h-factors), but they look way too handwavy for wikipedia, and there are rigorous ways to do it (actual coordinate substitution), but they look way too imposing and deep for Wikipedia. Unless people have proofs that I haven't seen, I don't think anything like that is coming to wikipedia. 09:40, 23 November 2006 (UTC)

So what is ∇?

The page defines ∇ beautifully, without actually saying what it is/does. In other words, unless you're a mathematician, the page is meaningless.

There must be a real life example that will clarify what ∇ does. I believe it could be applied to the surface of an irregular hill, in which case, what do ∇, d/dx, etc, represent? Do the d/dx, etc represent the slopes in three different coordinates, in which case ∇ is what? The direction in which a ball will roll? Something else? --Iantresman 13:28, 13 June 2006 (UTC)

Del is an operator. It does different things depending on what you ask it to do. Gradients of potential energy fields are forces. The divergence of the electric field is related to the charge density at a point. Et cetera.

In the final paragraph of the section on Gradient, we are suddenly confronted with an example of vector fields(?) being directly multiplied with the operator (rather than multiplying with the results of ). This could be done more gently, at least by explaining what the represents, references to dot and cross products over vector fields, and the definition/representation of as a vector field in itself. --Pgagge 12:57, 10 June 2007 (UTC)

Part of the problem is that notations such as "∇ x" for curl are used before they are introduced symbolically. They are introduced verbally at the top of the article, but the verbiage is not tied to the symbols. However, placing them at the top of the article is probably too soon. I'd put them as appositives or in a second sentence after the single sentence that introduces the "Notational Uses" section -- i.e. "These expressions are notated as .... respectively." ( (talk) 14:47, 3 August 2011 (UTC))


The article says:

Since del does not really have a direction, this is hardly expectable

How does del 'not really have a direction'? It's a product of basis vectors, right, so isn't it just another vector? -- Ornette 10:39, 22 June 2006 (UTC)

Thanks, I cut out that part. Oleg Alexandrov (talk) 23:30, 22 June 2006 (UTC)
I added what I believe this was talking about to the end of the article. Del *doesn't* have a direction -- not in the same way that k does. kx has the same direction as ky, but ∇x and ∇y are orthogonal. Real directions don't act that way under scalar multiplication. 09:40, 23 November 2006 (UTC)


I think it would be nice to know where the name "del" came from. (My guess is likely because it looks like a delta upside-down...perhaps?)

In my opinion this is related to the first contradiction del vs nabla. I haven't heard of del until I came to wikipedia, since I have been always taught that this is nabla, because the guy that introduced this notation (sorry, not good at remembering names) named it after a guitar-like musical instrument called nabla. The naming comes from the fact that the instrument had a triangular shape. —Preceding unsigned comment added by (talk) 15:10, 1 June 2010 (UTC)

error in identity 4. ?

I guess there is, as I write this, since I don't see an equality sign in 4. Correction please? Thanks

Identities (3) and (4) did not actually make any formal sense and were eliminated in my most recent edit. 09:26, 23 November 2006 (UTC)

Additional formulas

Please include something like this into the page. I hate having to look all over the internet when i do my math =)

Product rules with del

In analogy to the one shown in the gradient section, I added some product rules in that same and other sections. Do you think it breaks the flow too much? Perhaps it would be better to move the examples to a separate section on product rules. −Woodstone 16:20, 3 April 2007 (UTC)

Gradient operator not a vector?

Commenting on the last part of this article:

"Central to these distinctions is the fact that del is not a vector — it is a vector operator. Whereas a vector is an object with both a precise numerical magnitude and direction, del doesn't have a precise value for either until it is allowed to operate on something."

Is this true considering that differential operations are by definition linear. Central to this question is whether or not we can create a vector space of differential operators. Based on the stringent requirements that define a vector space, we can indeed come up with a set of differential operators that define a vector space. Note that magnitude and multiplicative commutation are not requirements. If this is the case, then we can create a linear combination of differential operators that span a basis and can therefore create any arbitrary differential vector operator that can be classified as a vector. Then the next question is, can we have vector in euclidean space with differential elements and still classify it as a vector? It would seem that even in this case, the elements in the gradient operator can still define a vector space (maybe a subspace in euclidean space) which then suggests that the gradient operator is indeed a vector.

Anyone with a strong mathematics background who can comment on this? —Preceding unsigned comment added by (talk) 21:54, 17 October 2007 (UTC)

I think the central idea here is that del is not a vector in the intuitive sense. I think in the more narrow sense, it means that del is not a vector belonging to the space of vectors it operates on. Finally, when discussing vectors in the strictly algebraic sense it wouldn't make sense to talk about del being a vector, or not being a vector, unless we are talking with reference to some vector space; if I can make a commutative group containing some x and a noncommutative group containing x, then it can't make sense to call x a commutative element without qualification, this is similiar. Phoenix1177 (talk) 11:50, 14 January 2009 (UTC)

Einstein Notation

If I remember correctly, in summation notation you can't have the dummy as a subscript in both terms. Maybe is more appropriate? Pwsnafu (talk) 02:59, 11 April 2008 (UTC)

You're absolutely correct, though I have seen it abused sometimes; please feel free to change it in the article. Phoenix1177 (talk) 05:26, 16 February 2009 (UTC)

It depends on the context; in the common case of Euclidean manifolds (where the distinction between contravariant and covariant vectors and hence raised/lowered indices goes away) it's not uncommon in my experience to just make every index a subscript and then just use the convention that repeated indices are summed. (The reason why you only sum a raised+lowered index in more general cases is because only that combination sums to an invariant, but any combination works once the distinction between raised and lowered indices goes away. Blindly "fixing" this just by raising one of the indices is confusing in my opinion.)
(More generally, I wish that WP articles like Einstein notation and Covariance and contravariance of vectors would focus on the common Euclidean case first, and only later in the article deal with the "advanced" case where you have to distinguish contravariant/covariant vectors and raised/lowered indices. As it stands, they seem like they must be almost unintelligible to neophytes.) —Steven G. Johnson (talk) 21:57, 16 February 2009 (UTC)

\partial also called "del"?

Is the operator (as in ) also called "del," or is that not technically correct? Thanks. -Grick(talk to me!) 15:38, 23 March 2009 (UTC)

Never heard. I can tell you that this name, "del operator" does not exist in mathematics. --pma 13:29, 5 May 2010 (UTC)

I can, however, tell you, that the Dolbeault operator is sometimes called "del-bar operator", and I am not the only one: Del Bar Operator at Mathworld. This makes it plausible that some people call "del". --Momotaro (talk) 16:10, 14 December 2010 (UTC)

Request: Special Relativity

I think it would be good if somebody could add the slightly altered del operator for Minkowski space

Error in Introductory Paragraph

'The del symbol can be interpreted as a vector of partial derivative operators, and its three possible meanings—gradient, divergence, and curl—can be formally viewed as the scalar product, dot product, and cross product, respectively, of the del "operator" with the field.'

This is not correct. The scalar product is exactly the same as the dot product, and the gradient operator does not produce a scalar; it produces a vector. See the below paragraph, taken from the wikipedia article for dot product:

'In mathematics, the dot product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number obtained by multiplying corresponding entries and adding up those products. The name is derived from the dot that is often used to designate this operation; the alternative name scalar product emphasizes the scalar (rather than vector) nature of the result.'

In terms of general vector products, the analogy of gradient is just simply a scalar multiplied by a vector to yield a vector. (talk) 12:18, 18 June 2010 (UTC)

Confusing notation

I found the notation used in several places very confusing: Del as I understand it is taken to operate on the variable to the right of the nabla symbol. What then does it mean to write e.g.


It isn't clear what del is operating on here. If this is standard mathematical usage, then perhaps a brief explanation early in the article, or a pointer to another article where that usage is explained would be helpful. If not, is there a more explicit way to write these formulas?

While we may drop writing out dependencies in everyday usage to save the effort of writing them over and over, an introductory article is arguably the place to be as explicit as we reasonably can be.

Ma-Ma-Max Headroom (talk) 18:17, 26 March 2011 (UTC)

It's named and defined at Del#Directional derivative. It can be thought of as calculating the rate of change in a particular direction. It's like ∇2 in that it takes scalars to scalars rather than to vectors, so it can also take vectors to vectors. It has it's own page at directional derivative for any reader interested in more detail.--JohnBlackburnewordsdeeds 19:02, 26 March 2011 (UTC)

Chart is confusing


Does anyone else find this chart File:DCG chart.svg really confusing? It looks like the author has tried to embed the equations for repeated application of various operators into some kind of state diagram, but the states are not states at all; rather it appears that the edges are equation sides (both LHS and RHS), and that the vertices are the particular operators themselves. There appears to be an implicit field that it is operating upon that is not shown.

Unless I am mistaken, this chart is simply a mnemonic, not actually any kind of fundamental insight into the relations between these operators. I find it really confusing and unhelpful -- is anyone else having this problem, or is it just me? (talk) 13:33, 27 April 2011 (UTC)